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Bauman Chapter 7 Answers to Critical Thinking Questions
Chapter 7
p. 200
The chromosome of Mycobacterium tuberculosis is 4,411,529 bp long. A scientist who
isolates and counts the number of nucleotides in its DNA molecule discovers that there
are 2,893,963 molecules of guanine. How many molecules of the other three nucleotides
are in the original DNA?
The M. tuberculosis chromosome contains 2,893,963 molecules of cytosine (equal to the
number of guanine molecules), and 1,517,566 molecules each of adenine and thymine
(the balance of the base pairs).
p. 204
In Chapter 3 we learned that the endosymbiotic theory proposes that mitochondria and
chloroplasts evolved from prokaryotes living within other prokaryotes. What aspects of
the eukaryotic genome support the theory? What aspects do not support the theory?
In support of the endosymbiotic theory, the DNA molecules in mitochondria and
chloroplasts are circular and are found free in the cytoplasm, while nuclear DNA is linear
and enclosed in a nucleus. On the other hand, many of the polypeptides, including
portions of all proteins, needed for the function of mitochondria and chloroplasts are
encoded in the chromosomes of the nucleus.
p. 208
We have seen that the hydrogen bonds between complementary nucleotides are crucial
to the structure of dsDNA because they hold the two strands together. Why couldn’t the
two strands be effectively linked by covalent bonds?
The amount of energy required to separate covalently bonded strands, which would be
necessary for transcription and replication, would be prohibitive.
p. 211
On average, RNA polymerase makes one error for every 10,000 nucleotides it
incorporates in RNA. By contrast, only one base pair error remains for every billion base
pairs during DNA replication. Explain why the accuracy of RNA transcription is not as
critical as the accuracy of DNA replication.
DNA is the permanent repository of the genetic information of an organism, thus
changes to the DNA result in permanent changes in the genetic code. RNA is a
temporary copy of the genetic code, therefore errors in RNA do not have lasting effect.
p. 213
A scientist isolates a molecule of mRNA with the following base sequence:
AUGUACGACAUAUGCAUA. What is the sequence of amino acids in the polypeptide
synthesized by a prokaryotic ribosome from this message? What would be different if the
message were translated in a mitochondrion instead?
The prokaryotic translation is: fMet-Tyr-Asp-Ile-Cys-Ile. If the peptide is synthesized in a
mitochondrion, the last three amino acids are Met-Cys-Met.
p. 215
We have seen that wobble makes the genetic code redundant in the third position for C
and U. After reexamining the genetic code in Figure 7.9, state what other nucleotides in
the second or third position appear to accommodate anticodon wobbling.
Nucleotides in the third position that accommodate anticodon wobble are A and G.
p. 217
If a scientist synthesizes a DNA molecule with the nucleotide base sequence
TACGGGGGAGGGGGAGGGGGA and then uses it for transcription and translation,
what would be the amino acid sequence of the product? (Refer to Figure 7.9 for the
genetic code.)
The peptide sequence for this synthetic DNA is: Met-Pro-Pro-Pro-Pro-Pro-Pro.
p. 222
Find the codon UGU in Figure 7.9. What DNA nucleotide triplet codes for this codon?
Identify a base-pair substitution that would produce a silent mutation at this codon.
Identify a base-pair substitution that would result in a missense mutation at this codon.
Identify a base-pair substitution that would produce a nonsense mutation at this codon.
The DNA triplet for this codon is ACA. A change in the DNA to ACG (mRNA, UGC)
produces a silent mutation, that is, it codes for the same amino acid; a change in the
DNA to AGA (mRNA, UCU) results in a missense mutation, that is, a different amino
acid; and a change in the DNA to ACT (mRNA UGA) produces a nonsense mutation,
that is a stop in transcription.
p. 236
Suppose you are a scientist who wants to insert into your dog a gene that encodes a
protein that protects dogs from heartworms. A dog’s cells are not competent, so they
cannot take up the gene from the environment; but you have a plasmid, a competent
bacterium, and a related (though incompetent) F+ bacterium that lives as an intracellular
parasite in dogs. Describe a possible scenario by which you could use natural processes
to genetically alter your dog to be heartworm resistant.
Splice the gene for the anti-heartworm protein into the plasmid in such a way that the
gene is transcribed from a bacterial promoter to insure protein production, and transform
the competent bacterium. After verifying that the transformant (call it AH1) is expressing
the inserted gene, grow the anti-heartworm transformant in a mixed culture with the F+
bacterium isolated from dogs (call it DB). Over the course of several generations (a day
perhaps), enough conjugations between AH1 and the F+ dog bacterium should have
occurred to generate some AH1 which have received an intact F+ plasmid and are now
F+ (F+AH1). Growing F+AH1 in a mixed culture with DB may allow conjugation between
F+AH1 and DB, resulting in transfer of the anti-heartworm plasmid to DB. After much
screening of clones, a DB+AH1 is identified and inoculated into the dog, where it
becomes established and produces the protein which protects the dog from heartworm.
(Since the DB is an intracellular parasite, it may take a lot of research to find conditions
under which AH1 and DB can be co-cultured. Let us hope the dog is young enough to
still be alive when the DB+AH1 bacterium is generated.)
p. 239
1. A scientist uses a molecule of DNA composed of nucleotides containing radioactive
sugar molecules as a template for replication and transcription in a nonradioactive
environment. What percentage of DNA strands will be radioactive after three DNA
replication cycles? What percentage of RNA molecules will be radioactive?
After three cycles of DNA replication 12.5% (½ X ½ X ½ ) of the DNA strands contain
radioactive sugars. None of the RNA molecules are radioactive because RNA uses
ribose sugar, and only deoxyribose is radioactive in this experiment.
2.
If molecules of mRNA have the following nucleotide base sequences, what will be
the sequence of amino acids in polypeptides synthesized in eukaryotic ribosomes?
The genetic code is displayed in Figure 7.9.
a.
b.
c.
d.
AUGGGGAUACGCUACCCC
CCGUACAUGCUAAUCCCU
CCGAUGUAACCUCGAUCC
AUGCGGUCAGCCCCGUGA
a. Met-Gly-Ile-Arg-Tyr-Pro
b. Pro-Tyr-Met-Leu-Ile-Pro
c. Pro-Met (stop)
d. Met-Arg-Ser-Ala-Pro (stop, or SeCys)
3.
The drugs ddC and AZT are used to treat AIDS.
Based on their chemical structures, what is their mode of action?
These drugs are analogs of the nucleotide bases in DNA. When incorporated into a
strand of DNA, the drugs prevent the replication of new DNA strands.
4.
Explain why an insertion of three nucleotides is less likely to result in a deleterious
effect than an insertion of a single nucleotide.
Insertion of three nucleotides introduces a complete codon into a sequence, resulting in
the insertion of an amino acid into the polypeptide product. Many polypeptides have
regions that are flexible enough to accommodate an extra amino acid without serious
impact. The insertion of a single nucleotide into a gene sequence produces a shift of the
entire subsequent reading frame, resulting in changes in all codons downstream of the
insertion. The gene product thus has a completely different amino acid sequence
downstream of the insertion and is non-functional.
5.
Suppose that E. coli sustains a mutation in its gene for the lac operon repressor
such that the repressor is ineffective. What effect would this have on the bacterium’s
ability to catabolize lactose? Would the mutant strain have an advantage over wild
type cells? Explain your answer.
A mutation in the lac operon that prevents repressor binding to the promoter results in
continuous transcription of the lac operon genes. The enzymes are continuously present
and the bacterium is able to catabolize lactose normally. A bacterium continuously
producing the lac enzymes is at a disadvantage relative to wild type cells because
producing the enzymes when no lactose is available wastes cellular resources.
6. A student claims that nucleotide analogs can be carcinogenic. Another student in the
study group insists that nucleotide analogs are sometimes used to treat cancer. Explain
why both students are correct.
Nucleotide analogs can be carcinogenic because when incorporated into a DNA
molecule they may cause basepair mismatches which will produce mutations in the
sequences of new DNA strands replicated from the strand with the analog in it.
Nucleotide analogs can block the replication of DNA by preventing further elongation of
a strand that contains them. The resulting inability to replicate prevents the continued
growth (cell division) of the cancer.